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Feb 16, 2021 at 22:41 vote accept Jarek Kędra
Mar 11, 2020 at 19:45 answer added J D timeline score: 0
Jul 12, 2016 at 12:43 comment added YCor It's unclear that this result has a precise author, because I guess that the finite-dimensional representations of the corresponding Lie algebra were classified (middle 19th? Clebsch?) much before the universal covering of $\mathrm{SL}_2(\mathbf{R})$ was defined (no idea where it appeared first!), and the only ingredient of the proof beyond basic Lie theory is this classification of representations of the Lie algebra.
Jul 12, 2016 at 12:38 history edited Jarek Kędra CC BY-SA 3.0
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Jul 1, 2016 at 7:17 comment added Jarek Kędra Thank you. Well, of course, it is not so difficult to give a proof. I thought that such a well known (and not obvious) fact should be proven in a textbook.
Jun 29, 2016 at 13:21 answer added Jeffrey Adams timeline score: 10
Jun 29, 2016 at 12:12 comment added Sean Lawton This is addressed here and here.
Jun 29, 2016 at 12:11 comment added Venkataramana Also see the link: mathoverflow.net/questions/110208/…
Jun 29, 2016 at 12:08 comment added Venkataramana It is easier to give a proof; every linear representation from the universal cover of $SL_2(\mathbb {R})$ into $GL_n(\mathbb {R})$ yields a complex representation of Lie algebras $\mathfrak {sl}_2(\mathbb {C})$ into $\mathfrak{gl}_n(\mathbb {C})$; since the group $SL_2(\mathbb {C})$ is simply connected, this extends to a holomorphic (algebraic) representation $SL_2(\mathbb {C}) \rightarrow GL_n(\mathbb {C}$; restricting , we get a representation $SL_2(\mathbb{R}\rightarrow GL_n(\mathbb {R})$. Hence the representation can never be faithful on any finite cover of $SL_2(\mathbb{R})$.
Jun 29, 2016 at 11:54 history asked Jarek Kędra CC BY-SA 3.0